Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{z^2 - 64}{z + 8}$
Answer: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = z$ $ b = \sqrt{64} = 8$ So we can rewrite the expression as: $x = \dfrac{({z} + {8})({z} {-8})} {z + 8} $ We can divide the numerator and denominator by $(z + 8)$ on condition that $z \neq -8$ Therefore $x = z - 8; z \neq -8$